Astudent constructs a frequency distribution where the frequency of the first class is 8 and a corresponding histogram with the area of the bar in the first class is 80. the frequency for the second class is twice the frequency of the first class. in the histogram, the area of the bar corresponding to the second class is four times the area of the bar corresponding to the first class. what is the frequency of the second class and the area of the bar corresponding to the second class? 16, 320 8, 320 16, 160 8, 160

The correct option is 1.

Step-by-step explanation:

The frequency of the first class = 8

The area of the bar in the first class = 80

It is given that the frequency for the second class is twice the frequency of the first class.

The frequency of the second class = 8 × 2 = 16

It is given that the area of the bar corresponding to the second class is four times the area of the bar corresponding to the first class.

The area of the bar in the second class = 80 × 4 = 320

The frequency of the second class and the area of the bar corresponding to the second class are 16 and 320 respectively.

Therefore, the correct option is 1.

sum= when you add is how much the whole amount is

distance= when you use angles and it means how far it goes

step-by-step explanation: